Question

obtain all the zerosif the polynomial x4-3x3-x2+9x-6,if two of its zeros are ^3 and -^3

Answer 1

Since √3 and -√3 are zeros , x-√3 and x+√3 are factors of the polynomial x^4-3x³-x²+9x-6 .

(x+√3)(x-√3) = x²-(√3)² [ (a+b)(a-b)=a²-b²]
                     = x²- 3

Now divide the polynomial x^4-3x³-x²+9x-6 by (x²-3)

=>  
                 x² - 3x + 2
              -----------------------
      x²-3 ║ x^4-3x³-x²+9x-6 
                 x^4      -3x²
                -           +
              --------------------------
                    0-3x³+2x²+9x-6
                      -3x³       +9x
                      +            -
             ---------------------------
                         0+2x²+0-6
                             2x²    -6
                            -         +
             ---------------------------
                             0       0
           

=>x^4-3x³-x²+9x-6 =  (x²-3x+2)(x²- 3)
 
=>(x²-3x+2) = x²-2x-x+2)
                    =x(x-2)-1(x-2)
                    =(x-2)(x-1)
                   =>x=2,1

=>x^4-3x³-x²+9x-6 = (x-2)(x-1)(x+√3)(x-√3) 

Therefore the zeros are 2,1,√3 and -√3
       

Answer 2

Answer:

Step-by-step explanation:

Zeroes of the polynomial given are -√3 & √3 . Therefore (x-√3) and (x+√3) are factor of p(x).

Multiply both the factors, which inturn is a factor of p(x).

After multiplying :- we get (x^2-3)

Now divide p(x) i.e (x^4-3x^3-x^2+9x-6) by (x^2-3)

After dividing p(x) by its factor i.e (x^2-3) we get qoutient as (x^2-3x+2) & remainder as 0.

Now we have to factorise the quotient by splitting the middle term method i.e

X^2-3x+2. Sum= -3

Product =2

-1 , -2

i.e (x+1)&(x+2) are factor of p(x)

i.e zeroes of the polynomial are -√3,√3,-1 and -2.